19 σ = P/A o σ B Maximum tensile strength σ % 0.2% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional

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19 σ = P/A o σ B Maximum tensile strength σ 0. 0.% 0.% proof stress σ EL Elastic limit Work hardening coefficient failure necking σ PL Proportional limit ε p = 0.% ε e = σ 0. /E plastic strain ε = ε e + ε p = σ 0. /E+0.00 ε = l/l o Uniform strain strain of failure elastic strain elastic deformation linear elasticity σ = Eε e E: FEM

nonlinear elasticity τ τ δ σ σ =σ o sinωt, ε = ε o sin(ωt δ) W = σdε = σ dε dt 1cycle dt t dε dt =ε oωcos(ωt δ) W = σ o ε o ω sinωt cos(ωt δ) dt 1cycle = σ o ε o (sinωt cosωt cosδ + sin ωt sinδ)d (ωt) = σ o ε o 0 π π sinδ W = σ o ε o / W W = π sinδ πδ W ε

A(t)/A o A t n+1 A = A o cosωt exp( αωt ) t n n n+1 ωt n = nπ A n A n+1 = exp{αω (t n +1 t n )}= exp(απ ) α 1 π ln A n = α A n+1 α = δ A n A n+1 t

plastic deformation crystal amorphous d τ x τ = Gγ = G x d G γ = x d b d x τ τ

b τ atom = τ max sin πx b x << b τ atom πτ max b x τ max = G π b d {111} <110> d {111} b <110> τ atom d = a 3 b = a b d = 3 0.61 τ max 0 b/ b x τ max G 10 GPa 1/10 1/100 TTT

dislocation σ y edge dislocation screw b b b edg dislocation screw dislocation b b {111}/<110>

edge screw deformation by twinnig twin mechanical twin τ τ b τ τ

DBTT annealing twin Ni τ l x τ b V = lxb W = τv = τlbx l x F W = Fx τ F = τlb f

f = F l = τb r σ ij = Gb f ij (θ ) πk r K = 1 K = 1 ν ν f ij (θ) θ σ θz = σ zθ = Gb πr ε θ z = ε zθ = b 4πr R strain energy E e screw = R σ θ z ε θ z + σ ε zθ zθ πrdr = G b r o 4π ln R r o r o E edge e = G b 4πK ln R r o R E e G b line tension l l + l E e = E e (l + l ) E e l = E e l

T T l W = T l T = E Gb t l F τ R θ b l F = τbl l Tsinθ Tsinθ θ θ Tsinθ T T R F = T sinθ τ τ = T bl Gb sinθ = sinθ l sinθ θ = π / sinθ =1 forest dislocation τ = Gb sinθ l τ = Gb l ductile material

σ i σ ob σ dis σ gb σ =σ i +σ ob + σ dis + σ gb dislocation density l dis ρ l dis = 1 ρ D σ gb = k D k σ =σ i +α Gb l ob + βgb ρ + k D l ob α β σ o σ =σ o + k D 100 nm

brittle materials Al O 3 ZrO SiC 3 N 4 + + + + + + + + + + b + + + b b + + d x λ σ atom = σ max sin πx λ x << λ πx σ atom σ max λ σ atom σ ε d σ = Eε = E x d E B x d λ/ σ max = E π λ d σ max λ d

σ max E π E 6 x = 0 ~ λ/ σ atom 0 λ/ E B λ λ / E B = σ atom dx = σ 0 max sin πx dx = σ λ max 0 λ π γ s γ s = E B σ max λ π = γ s σ max λ σ max = Eγ s d σ t C C σ x y σ y σ yy x σ yy = πx = σ πc I

I d σ yy σ yy = πd Eγ s d σ f K =σ f πc σ f Eγ s C t u e = σ E S S = αc α U e = u e St = α σ E C t U e, W s, U A A = Ct W s = Aγ s = 4Ctγ s U max U U = U e +W s = α σ E C t+4ctγ s C* C C* U U max U e U C > C* W s

U U A = 1 U t C = ασ E C+γ =0 s σ f = 4Eγ s αc α = π σ f = Eγ s πc C C = Eγ s C fracture toughness = Yσ πc Y crack geometry factor Y = 1.1 G I, Γ IC G = U e A G IC = γ s elastic energy release rate I A = Ct U e = α σ E C t = σ πc t = σ πa A E 4Et G I = U e A = σ π Et A = σ πc E G I

G I G IC G IC = W s A = γ s IC critical elastic energy release rate fracture energy IC = σ πc G I = σ πc E G I = K I E C = EG IC G IC = K IC E U = U e +W s +W p W p U A = U e A + (W +W ) s p = 0 A γ p U = πσ C t + 4γ E s Ct + γ p Ct U A = U t C = πσ C E + γ s + γ p = 0

σ πc = E(γ s + γ p ) C = E(γ s + γ p ) G IC = γ s + γ p C = 16 MPa m 0.5 γ s =.8 J/m E = 411 GPa G IC = K IC E = (16 10 6 ) = 63 J/m 411 10 9 γ p = G IC γ s = 617 J/m C Eγ p G IC γ p C r y y = r Y = 1 πr y π C r y = σ πc σ yy x r Y C = 1 σ

y y' r p x = 0 r y x = r y r p σ' yy C r y πx σ r p 0 Y dx = σ r Y dx y σ yy r y σ' yy r p x r p = r y 0 πx dx = π r y = K I r πσ p = 1 Y π = r y r p r y σ yy r y effective crack length C eff = (C + r y ) = C 1+ r y = C 1+ 1 σ C effective stress intensity factor K eff =σ πc eff =σ πc 1+ 1 σ = 1+ 1 σ